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1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise
3. Ex 5.1

Transcript

Ex 5.1, 11 Find all points of discontinuity of f, where f is defined by π(π₯)={ β(π₯3β3, ππ π₯β€2@&π₯2+1 , ππ π₯>2)β€ Since we need to find continuity at of the function We check continuity for different values of x When x < 2 When x = 2 When x > 2 Case 1 : When x < 2 For x < 2, f(x) = π₯^3β3 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 3 Case 2 : When x = 2 f(x) is continuous at π₯ =2 if L.H.L = R.H.L = π(2) if limβ¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) " " π(π₯)= π(2) Since there are two different functions on the left & right of 2, we take LHL & RHL . LHL at x β 2 limβ¬(xβ2^β ) f(x) = limβ¬(hβ0) f(2 β h) = limβ¬(hβ0) ((2ββ)^3β3) = (2β0)^3β3 = 2^3β3 = 8 β 3 = 5 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) (2+β)^2+1 = (2 + 0)2 + 1 = 22 + 1 = 4 + 1 = 5 & π(2) = π₯^3β3 = 2^3β3 = 8 β 3 = 5 Hence, L.H.L = R.H.L = π(2) β΄ f is continuous at x=2 Case 3 : When x > 2 For x > 2, f(x) = x2 + 1 Since this a polynomial It is continuous β΄ f(x) is continuous for x > 2 Hence, there is no point of discontinuity Thus, f is continuous for all πβπ

Ex 5.1